Question

A battery of 9v is connected in a series with resistor of 0.2Ω ,0.3Ω,0.4Ω,0.5Ω and 12Ω respectively.how much current would flow through the 12Ω resistor?​

Answer 1

${\bold{\pink{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}$

All the resistor are connected in series.so,

• Combined resistance (R) = 0.2 + 0.3 + 0.4 +0.5 + 12

• Combined resistance (R) =13.4Ω

• Potential difference (V) = 9 v

${\bold{\pink{\underline{\pink{To}\:\purple{Fin}\blue{d}\red{:-}}}}}$

• Current (I) = ?

${\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}$

Current flow through the whole circuit can be calculated by using ohm's law:-

$\bf\:\dfrac{V}{I}=R$

$\bf\:\dfrac{9}{I}=13.4$

$\bf\:I=\dfrac{9}{13.4}$

$\bf\:I=0.67\:ampere$

Since the same current flows through all the resistors connected in series in a circuit,therefore 0.67 ampere of current would flow through the 12Ω resistor.